Sky Pond Incubation Experiment (at USGS Fort)

Nutrient Uptake/Release Data

Data Visualization - TDN uptake

Before we jump into any statistical analyses, just want to get an idea of how things play out by treatment and temperature .

TDN uptake - Data table summary

Incubation experiment nutrient data summary: means, standard deviation, and number of replicates per treatment
tempxtrt treatment temperature meanchangeTDN_uM sdchangeTDN_uM n
8Control C 8 7.204 2.053 6
8N+P NP 8 48.590 3.815 6
8Nitrogen N 8 14.838 3.488 6
8Phosphorus P 8 9.863 0.542 5
12Control C 12 10.255 0.826 6
12N+P NP 12 58.205 1.072 6
12Nitrogen N 12 23.563 5.616 6
12Phosphorus P 12 10.051 0.805 6
16Control C 16 8.045 1.400 6
16N+P NP 16 54.058 1.616 6
16Nitrogen N 16 43.418 3.416 6
16Phosphorus P 16 8.834 1.614 6

TDN uptake - Descriptive statistics and plots

Here we have some summary statistics for TDN uptake. First we are looking at change in TDN from start to end of experiment by TEMPERTURE and by TREATMENT separately. I’ve also plotted boxplots with individual points.

## ------------------------------------------------------------------------- 
## changeTDN_uM ~ temperature
## 
## Summary: 
## n pairs: 71, valid: 71 (100.0%), missings: 0 (0.0%), groups: 3
## 
##                            
##             8     12     16
## mean     20.6   25.5   28.6
## median   10.5   14.1   25.0
## sd       17.5   20.3   21.0
## IQR      22.0   27.0   40.1
## n          23     24     24
## np      32.4%  33.8%  33.8%
## NAs         0      0      0
## 0s          0      0      0
## 
## Kruskal-Wallis rank sum test:
##   Kruskal-Wallis chi-squared = 2.5284, df = 2, p-value = 0.2825

## ------------------------------------------------------------------------- 
## changeTDN_uM ~ treatment
## 
## Summary: 
## n pairs: 71, valid: 71 (100.0%), missings: 0 (0.0%), groups: 4
## 
##                                   
##             C      N      P     NP
## mean      8.5   27.3    9.6   53.6
## median    8.8   21.3    9.9   53.8
## sd        1.9   13.0    1.2    4.7
## IQR       3.6   22.7    1.5    5.7
## n          18     18     17     18
## np      25.4%  25.4%  23.9%  25.4%
## NAs         0      0      0      0
## 0s          0      0      0      0
## 
## Kruskal-Wallis rank sum test:
##   Kruskal-Wallis chi-squared = 58.424, df = 3, p-value = 1.276e-12

TDN uptake - ANOVA results - with intxn

From our preliminary glance at the data, it appears there is an interaction between the temperature of the experiment and the nutrient treatment. In the N-only treatment, TDN uptake increased with temperature. Interesting, in the N+P nutrient treatment, almost all of the TDN was consumed in the experiment. The following two plots are interaction plots of by nutrient treatment and temperature treatment, respecitvely.

## Anova Table (Type III tests)
## 
## Response: changeTDN_uM
##                       Sum Sq Df F value    Pr(>F)    
## (Intercept)            311.4  1  43.721 1.222e-08 ***
## temperature             29.8  2   2.093    0.1324    
## treatment             6553.0  3 306.723 < 2.2e-16 ***
## temperature:treatment 1980.6  6  46.354 < 2.2e-16 ***
## Residuals              420.2 59                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Studying the output of the ANOVA table (Type III SS), we see there is a evidence of a significant interaction effect (F=46.453, p<0.0001). The test for the main effect of nutrient treatment is significant (p<0.0001) but the test for the main effect of temperature alone is not significant (p=0.13). After fitting an ANOVA model it is important to always check the relevant model assumptions. This includes making QQ-plots and residual plots. Will use Levene’s test of equality of variances too.

## Levene's Test for Homogeneity of Variance (center = median)
##       Df F value  Pr(>F)  
## group 11  1.8448 0.06639 .
##       59                  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Neither plot indicates a significant violation of normality assumptions. We’re good here.

TDN uptake - Pairwise comparisons

Pairwise comparsions for each level of temperature

## $emmeans
## temperature = 8:
##  treatment    emmean       SE df  lower.CL  upper.CL
##  C          7.203667 1.089457 59  5.023668  9.383665
##  N         14.838333 1.089457 59 12.658335 17.018332
##  P          9.863000 1.193440 59  7.474931 12.251069
##  NP        48.590000 1.089457 59 46.410002 50.769998
## 
## temperature = 12:
##  treatment    emmean       SE df  lower.CL  upper.CL
##  C         10.255333 1.089457 59  8.075335 12.435332
##  N         23.563333 1.089457 59 21.383335 25.743332
##  P         10.050667 1.089457 59  7.870668 12.230665
##  NP        58.205000 1.089457 59 56.025002 60.384998
## 
## temperature = 16:
##  treatment    emmean       SE df  lower.CL  upper.CL
##  C          8.045333 1.089457 59  5.865335 10.225332
##  N         43.418333 1.089457 59 41.238335 45.598332
##  P          8.834000 1.089457 59  6.654002 11.013998
##  NP        54.058333 1.089457 59 51.878335 56.238332
## 
## Confidence level used: 0.95 
## 
## $contrasts
## temperature = 8:
##  contrast    estimate       SE df t.ratio p.value
##  C - N     -7.6346667 1.540725 59  -4.955  <.0001
##  C - P     -2.6593333 1.615926 59  -1.646  0.3615
##  C - NP   -41.3863333 1.540725 59 -26.862  <.0001
##  N - P      4.9753333 1.615926 59   3.079  0.0162
##  N - NP   -33.7516667 1.540725 59 -21.906  <.0001
##  P - NP   -38.7270000 1.615926 59 -23.966  <.0001
## 
## temperature = 12:
##  contrast    estimate       SE df t.ratio p.value
##  C - N    -13.3080000 1.540725 59  -8.637  <.0001
##  C - P      0.2046667 1.540725 59   0.133  0.9992
##  C - NP   -47.9496667 1.540725 59 -31.121  <.0001
##  N - P     13.5126667 1.540725 59   8.770  <.0001
##  N - NP   -34.6416667 1.540725 59 -22.484  <.0001
##  P - NP   -48.1543333 1.540725 59 -31.254  <.0001
## 
## temperature = 16:
##  contrast    estimate       SE df t.ratio p.value
##  C - N    -35.3730000 1.540725 59 -22.959  <.0001
##  C - P     -0.7886667 1.540725 59  -0.512  0.9559
##  C - NP   -46.0130000 1.540725 59 -29.865  <.0001
##  N - P     34.5843333 1.540725 59  22.447  <.0001
##  N - NP   -10.6400000 1.540725 59  -6.906  <.0001
##  P - NP   -45.2243333 1.540725 59 -29.353  <.0001
## 
## P value adjustment: tukey method for comparing a family of 4 estimates

Pairwise comparsions for each level of nutrient treatment

## $emmeans
## treatment = C:
##  temperature    emmean       SE df  lower.CL  upper.CL
##  8            7.203667 1.089457 59  5.023668  9.383665
##  12          10.255333 1.089457 59  8.075335 12.435332
##  16           8.045333 1.089457 59  5.865335 10.225332
## 
## treatment = N:
##  temperature    emmean       SE df  lower.CL  upper.CL
##  8           14.838333 1.089457 59 12.658335 17.018332
##  12          23.563333 1.089457 59 21.383335 25.743332
##  16          43.418333 1.089457 59 41.238335 45.598332
## 
## treatment = P:
##  temperature    emmean       SE df  lower.CL  upper.CL
##  8            9.863000 1.193440 59  7.474931 12.251069
##  12          10.050667 1.089457 59  7.870668 12.230665
##  16           8.834000 1.089457 59  6.654002 11.013998
## 
## treatment = NP:
##  temperature    emmean       SE df  lower.CL  upper.CL
##  8           48.590000 1.089457 59 46.410002 50.769998
##  12          58.205000 1.089457 59 56.025002 60.384998
##  16          54.058333 1.089457 59 51.878335 56.238332
## 
## Confidence level used: 0.95 
## 
## $contrasts
## treatment = C:
##  contrast    estimate       SE df t.ratio p.value
##  8 - 12    -3.0516667 1.540725 59  -1.981  0.1259
##  8 - 16    -0.8416667 1.540725 59  -0.546  0.8488
##  12 - 16    2.2100000 1.540725 59   1.434  0.3301
## 
## treatment = N:
##  contrast    estimate       SE df t.ratio p.value
##  8 - 12    -8.7250000 1.540725 59  -5.663  <.0001
##  8 - 16   -28.5800000 1.540725 59 -18.550  <.0001
##  12 - 16  -19.8550000 1.540725 59 -12.887  <.0001
## 
## treatment = P:
##  contrast    estimate       SE df t.ratio p.value
##  8 - 12    -0.1876667 1.615926 59  -0.116  0.9926
##  8 - 16     1.0290000 1.615926 59   0.637  0.8005
##  12 - 16    1.2166667 1.540725 59   0.790  0.7108
## 
## treatment = NP:
##  contrast    estimate       SE df t.ratio p.value
##  8 - 12    -9.6150000 1.540725 59  -6.241  <.0001
##  8 - 16    -5.4683333 1.540725 59  -3.549  0.0022
##  12 - 16    4.1466667 1.540725 59   2.691  0.0247
## 
## P value adjustment: tukey method for comparing a family of 3 estimates
## NOTE: 'cld()' groupings are determined separately for each list member.
## $emmeans
##  temperature treatment    emmean       SE df  lower.CL  upper.CL .group 
##  8           C          7.203667 1.089457 59  5.023668  9.383665  |     
##  16          C          8.045333 1.089457 59  5.865335 10.225332  |     
##  16          P          8.834000 1.089457 59  6.654002 11.013998  ||    
##  8           P          9.863000 1.193440 59  7.474931 12.251069  ||    
##  12          P         10.050667 1.089457 59  7.870668 12.230665  ||    
##  12          C         10.255333 1.089457 59  8.075335 12.435332  ||    
##  8           N         14.838333 1.089457 59 12.658335 17.018332   |    
##  12          N         23.563333 1.089457 59 21.383335 25.743332    |   
##  16          N         43.418333 1.089457 59 41.238335 45.598332     |  
##  8           NP        48.590000 1.089457 59 46.410002 50.769998     || 
##  16          NP        54.058333 1.089457 59 51.878335 56.238332      ||
##  12          NP        58.205000 1.089457 59 56.025002 60.384998       |
## 
## Confidence level used: 0.95 
## P value adjustment: tukey method for comparing a family of 12 estimates 
## significance level used: alpha = 0.01 
## 
## $contrasts
##  contrast treatment    estimate       SE df t.ratio p.value .group 
##  8 - 16   N         -28.5800000 1.540725 59 -18.550  <.0001  |     
##  12 - 16  N         -19.8550000 1.540725 59 -12.887  <.0001   |    
##  8 - 12   NP         -9.6150000 1.540725 59  -6.241  <.0001    |   
##  8 - 12   N          -8.7250000 1.540725 59  -5.663  <.0001    ||  
##  8 - 16   NP         -5.4683333 1.540725 59  -3.549  0.0022    ||| 
##  8 - 12   C          -3.0516667 1.540725 59  -1.981  0.1259    ||||
##  8 - 16   C          -0.8416667 1.540725 59  -0.546  0.8488     |||
##  8 - 12   P          -0.1876667 1.615926 59  -0.116  0.9926     |||
##  8 - 16   P           1.0290000 1.615926 59   0.637  0.8005      ||
##  12 - 16  P           1.2166667 1.540725 59   0.790  0.7108      ||
##  12 - 16  C           2.2100000 1.540725 59   1.434  0.3301      ||
##  12 - 16  NP          4.1466667 1.540725 59   2.691  0.0247       |
## 
## P value adjustment: tukey method for comparing a family of 3 estimates 
## P value adjustment: tukey method for comparing a family of 12 estimates 
## significance level used: alpha = 0.01

Data Visualization - DOC exudation

Before we jump into any statistical analyses, just want to get an idea of how things play out by treatment and temperature .

DOC exudation - Data table summary

Incubation experiment nutrient data summary: means, standard deviation, and number of replicates per treatment
tempxtrt treatment temperature meanchangeDOC_mgL sdchangeDOC_mgL n
8Control C 8 -2.759 0.627 6
8N+P NP 8 -2.703 0.482 6
8Nitrogen N 8 -3.136 1.338 6
8Phosphorus P 8 -2.329 0.565 5
12Control C 12 -1.898 0.347 6
12N+P NP 12 -1.224 0.258 6
12Nitrogen N 12 -1.771 0.226 6
12Phosphorus P 12 -1.268 0.159 6
16Control C 16 -2.534 0.344 6
16N+P NP 16 -1.446 0.387 6
16Nitrogen N 16 -2.841 0.481 6
16Phosphorus P 16 -1.253 0.209 6

DOC exudation - Descriptive statistics and plots

Here we have some summary statistics for DOC exudation. First we are looking at change in DOC from start to end of experiment by TEMPERTURE and by TREATMENT separately. I’ve also plotted boxplots with individual points. Due to the way I calculated these values (Initial DOC minus final DOC), negative numbers here mean there is MORE dissolved organic carbon at the end of the experiment than the begining.

## ------------------------------------------------------------------------- 
## changeDOC_mgL ~ temperature
## 
## Summary: 
## n pairs: 71, valid: 71 (100.0%), missings: 0 (0.0%), groups: 3
## 
##                            
##             8     12     16
## mean     -2.7   -1.5   -2.0
## median   -2.8   -1.4   -2.0
## sd        0.8    0.4    0.8
## IQR       0.9    0.5    1.3
## n          23     24     24
## np      32.4%  33.8%  33.8%
## NAs         0      0      0
## 0s          0      0      0
## 
## Kruskal-Wallis rank sum test:
##   Kruskal-Wallis chi-squared = 24.495, df = 2, p-value = 4.796e-06

## ------------------------------------------------------------------------- 
## changeDOC_mgL ~ treatment
## 
## Summary: 
## n pairs: 71, valid: 71 (100.0%), missings: 0 (0.0%), groups: 4
## 
##                                   
##             C      N      P     NP
## mean     -2.4   -2.6   -1.6   -1.8
## median   -2.3   -2.2   -1.4   -1.6
## sd        0.6    1.0    0.6    0.8
## IQR       0.8    1.2    0.4    1.1
## n          18     18     17     18
## np      25.4%  25.4%  23.9%  25.4%
## NAs         0      0      0      0
## 0s          0      0      0      0
## 
## Kruskal-Wallis rank sum test:
##   Kruskal-Wallis chi-squared = 20.608, df = 3, p-value = 0.000127

DOC exudation - ANOVA results - with intxn

From our preliminary glance at the data, it looks like at lower temperatures, there is more DOC released (still pondering the ecological significance of that… fueling microbial loop?). At higher temperatures, it looks like there is less DOC at the end of the experiment.

## Anova Table (Type III tests)
## 
## Response: changeDOC_mgL
##                       Sum Sq Df  F value  Pr(>F)    
## (Intercept)           45.684  1 154.9900 < 2e-16 ***
## temperature            2.397  2   4.0659 0.02217 *  
## treatment              1.794  3   2.0290 0.11959    
## temperature:treatment  3.517  6   1.9887 0.08164 .  
## Residuals             17.390 59                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Studying the output of the ANOVA table (Type III SS), we see there is a evidence of a marginally significant interaction effect (F(6,59)=1.9887, p=0.08164). The test for the main effect of temperature treatment is significant (p=0.02217) but the test for the main effect of nutrient treatment alone is not significant (p=0.11959). After fitting an ANOVA model it is important to always check the relevant model assumptions. This includes making QQ-plots and residual plots. Will use Levene’s test of equality of variances too.

## Levene's Test for Homogeneity of Variance (center = median)
##       Df F value  Pr(>F)  
## group 11  2.0599 0.03806 *
##       59                  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

It looks like there is just one potential outlier in the data. Overall, neither plot indicates a significant violation of normality assumptions. We’re good here.

## Anova Table (Type III tests)
## 
## Response: changeDOC_mgL
##                        Sum Sq Df  F value    Pr(>F)    
## (Intercept)           15.2461  1 152.4365 4.749e-16 ***
## temperature            0.8119  2   4.0587 0.0239587 *  
## treatment              0.4805  3   1.6013 0.2023223    
## temperature:treatment  3.3104  6   5.5164 0.0002379 ***
## Residuals              4.5007 45                       
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

## Levene's Test for Homogeneity of Variance (center = median)
##       Df F value Pr(>F)
## group 11  0.7227 0.7109
##       45

If we take the one super high DOC value out, we meet the assumptions of normality. Not sure if this is really ethical though.

DOC exudation - Pairwise comparisons

Pairwise comparsions for each level of temperature. At 12 degrees, we see significnant differences between C - NP, N - P, and N - NP. At 16 degress, there is a difference between C - P, C-NP, N-P. There are no differences at 8 degrees.

## $emmeans
## temperature = 8:
##  treatment    emmean        SE df  lower.CL   upper.CL
##  C         -2.254333 0.1825886 45 -2.622086 -1.8865811
##  N         -1.841000 0.2236244 45 -2.291403 -1.3905973
##  P         -2.329000 0.1414325 45 -2.613860 -2.0441403
##  NP        -2.423500 0.1581263 45 -2.741983 -2.1050172
## 
## temperature = 12:
##  treatment    emmean        SE df  lower.CL   upper.CL
##  C         -1.897667 0.1291096 45 -2.157707 -1.6376266
##  N         -1.771000 0.1291096 45 -2.031040 -1.5109599
##  P         -1.267667 0.1291096 45 -1.527707 -1.0076266
##  NP        -1.224333 0.1291096 45 -1.484373 -0.9642932
## 
## temperature = 16:
##  treatment    emmean        SE df  lower.CL   upper.CL
##  C         -2.433000 0.1414325 45 -2.717860 -2.1481403
##  N         -2.231000 0.2236244 45 -2.681403 -1.7805973
##  P         -1.253333 0.1291096 45 -1.513373 -0.9932932
##  NP        -1.446000 0.1291096 45 -1.706040 -1.1859599
## 
## Confidence level used: 0.95 
## 
## $contrasts
## temperature = 8:
##  contrast    estimate        SE df t.ratio p.value
##  C - N    -0.41333333 0.2886979 45  -1.432  0.4867
##  C - P     0.07466667 0.2309583 45   0.323  0.9882
##  C - NP    0.16916667 0.2415420 45   0.700  0.8964
##  N - P     0.48800000 0.2645960 45   1.844  0.2666
##  N - NP    0.58250000 0.2738829 45   2.127  0.1602
##  P - NP    0.09450000 0.2121487 45   0.445  0.9702
## 
## temperature = 12:
##  contrast    estimate        SE df t.ratio p.value
##  C - N    -0.12666667 0.1825886 45  -0.694  0.8990
##  C - P    -0.63000000 0.1825886 45  -3.450  0.0065
##  C - NP   -0.67333333 0.1825886 45  -3.688  0.0033
##  N - P    -0.50333333 0.1825886 45  -2.757  0.0404
##  N - NP   -0.54666667 0.1825886 45  -2.994  0.0223
##  P - NP   -0.04333333 0.1825886 45  -0.237  0.9952
## 
## temperature = 16:
##  contrast    estimate        SE df t.ratio p.value
##  C - N    -0.20200000 0.2645960 45  -0.763  0.8704
##  C - P    -1.17966667 0.1915005 45  -6.160  <.0001
##  C - NP   -0.98700000 0.1915005 45  -5.154  <.0001
##  N - P    -0.97766667 0.2582192 45  -3.786  0.0025
##  N - NP   -0.78500000 0.2582192 45  -3.040  0.0198
##  P - NP    0.19266667 0.1825886 45   1.055  0.7181
## 
## P value adjustment: tukey method for comparing a family of 4 estimates

Pairwise comparsions for each level of nutrient treatment. In the control, there is a difference between 12-16 degrees. In the P treatment, there is a difference between 8-12, and 8-16. And in the NP treatment, there is a difference between 8-12, and 8-16.

## $emmeans
## treatment = C:
##  temperature    emmean        SE df  lower.CL   upper.CL
##  8           -2.254333 0.1825886 45 -2.622086 -1.8865811
##  12          -1.897667 0.1291096 45 -2.157707 -1.6376266
##  16          -2.433000 0.1414325 45 -2.717860 -2.1481403
## 
## treatment = N:
##  temperature    emmean        SE df  lower.CL   upper.CL
##  8           -1.841000 0.2236244 45 -2.291403 -1.3905973
##  12          -1.771000 0.1291096 45 -2.031040 -1.5109599
##  16          -2.231000 0.2236244 45 -2.681403 -1.7805973
## 
## treatment = P:
##  temperature    emmean        SE df  lower.CL   upper.CL
##  8           -2.329000 0.1414325 45 -2.613860 -2.0441403
##  12          -1.267667 0.1291096 45 -1.527707 -1.0076266
##  16          -1.253333 0.1291096 45 -1.513373 -0.9932932
## 
## treatment = NP:
##  temperature    emmean        SE df  lower.CL   upper.CL
##  8           -2.423500 0.1581263 45 -2.741983 -2.1050172
##  12          -1.224333 0.1291096 45 -1.484373 -0.9642932
##  16          -1.446000 0.1291096 45 -1.706040 -1.1859599
## 
## Confidence level used: 0.95 
## 
## $contrasts
## treatment = C:
##  contrast    estimate        SE df t.ratio p.value
##  8 - 12   -0.35666667 0.2236244 45  -1.595  0.2583
##  8 - 16    0.17866667 0.2309583 45   0.774  0.7210
##  12 - 16   0.53533333 0.1915005 45   2.795  0.0204
## 
## treatment = N:
##  contrast    estimate        SE df t.ratio p.value
##  8 - 12   -0.07000000 0.2582192 45  -0.271  0.9603
##  8 - 16    0.39000000 0.3162527 45   1.233  0.4402
##  12 - 16   0.46000000 0.2582192 45   1.781  0.1873
## 
## treatment = P:
##  contrast    estimate        SE df t.ratio p.value
##  8 - 12   -1.06133333 0.1915005 45  -5.542  <.0001
##  8 - 16   -1.07566667 0.1915005 45  -5.617  <.0001
##  12 - 16  -0.01433333 0.1825886 45  -0.079  0.9966
## 
## treatment = NP:
##  contrast    estimate        SE df t.ratio p.value
##  8 - 12   -1.19916667 0.2041402 45  -5.874  <.0001
##  8 - 16   -0.97750000 0.2041402 45  -4.788  0.0001
##  12 - 16   0.22166667 0.1825886 45   1.214  0.4512
## 
## P value adjustment: tukey method for comparing a family of 3 estimates

Data Visualization - PO4 uptake/release

Before we jump into any statistical analyses, just want to get an idea of how things play out by treatment and temperature .

PO4 uptake/release - Data table summary

Incubation experiment nutrient data summary: means, standard deviation, and number of replicates per treatment
tempxtrt treatment temperature meanchangePO4_uM sdchangePO4_uM n
8Control C 8 -5.455 2.146 6
8N+P NP 8 -1.083 1.103 6
8Nitrogen N 8 -8.886 1.358 6
8Phosphorus P 8 1.668 0.267 5
12Control C 12 -4.806 2.353 6
12N+P NP 12 7.893 0.000 6
12Nitrogen N 12 -5.016 4.002 6
12Phosphorus P 12 7.493 0.000 6
16Control C 16 -0.007 0.000 6
16N+P NP 16 7.893 0.000 6
16Nitrogen N 16 -0.007 0.000 6
16Phosphorus P 16 7.493 0.000 6

PO4 uptake/release - Descriptive statistics and plots

Here we have some summary statistics for PO4 uptake/release. First we are looking at change in PO4 from start to end of experiment by TEMPERTURE and by TREATMENT separately. I’ve also plotted boxplots with individual points. Due to the way I calculated these values (Initial PO4 minus final PO4), negative numbers here mean there is MORE phosphate at the end of the experiment than the begining. Positive number indiciate uptake… slightly counterintuitive.

## ------------------------------------------------------------------------- 
## changePO4_uM ~ temperature
## 
## Summary: 
## n pairs: 71, valid: 71 (100.0%), missings: 0 (0.0%), groups: 3
## 
##                            
##             8     12     16
## mean     -3.7    1.4    3.8
## median   -3.3    3.7    3.7
## sd        4.3    6.8    3.9
## IQR       6.9   13.3    7.6
## n          23     24     24
## np      32.4%  33.8%  33.8%
## NAs         0      0      0
## 0s          0      0      0
## 
## Kruskal-Wallis rank sum test:
##   Kruskal-Wallis chi-squared = 20.651, df = 2, p-value = 3.279e-05

## ------------------------------------------------------------------------- 
## changePO4_uM ~ treatment
## 
## Summary: 
## n pairs: 71, valid: 71 (100.0%), missings: 0 (0.0%), groups: 4
## 
##                                   
##             C      N      P     NP
## mean     -3.4   -4.6    5.8    4.9
## median   -5.7   -6.7    7.5    7.9
## sd        3.0    4.4    2.7    4.4
## IQR       5.9    8.4    5.5    8.3
## n          18     18     17     18
## np      25.4%  25.4%  23.9%  25.4%
## NAs         0      0      0      0
## 0s          0      0      0      0
## 
## Kruskal-Wallis rank sum test:
##   Kruskal-Wallis chi-squared = 39.501, df = 3, p-value = 1.359e-08

PO4 uptake/release - ANOVA results - with intxn

From our preliminary glance at the data, it looks like PO4 was released (averaging over all nutrient treatments). Interestingly, in the P & NP treatments, there was PO4 uptake compared to C & N treatments, where there was more PO4 leftover in the begining of the experiment than there were at the beginning.

## Anova Table (Type III tests)
## 
## Response: changePO4_uM
##                       Sum Sq Df F value    Pr(>F)    
## (Intercept)           178.54  1 71.9709 8.366e-12 ***
## temperature           106.27  2 21.4189 1.016e-07 ***
## treatment             364.96  3 49.0382 4.970e-16 ***
## temperature:treatment 143.09  6  9.6131 2.305e-07 ***
## Residuals             146.36 59                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Studying the output of the ANOVA table (Type III SS), we see there is a evidence of a marginally significant interaction effect (F(6,59)=9.6131, p<0.0001). The tests for the main effects of temperature and nutrient treatments are BOTH significant. After fitting an ANOVA model it is important to always check the relevant model assumptions. This includes making QQ-plots and residual plots. Will use Levene’s test of equality of variances too.

## Levene's Test for Homogeneity of Variance (center = median)
##       Df F value   Pr(>F)   
## group 11  2.9936 0.003176 **
##       59                    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

PO4 uptake does not pass the Levene’s test and the Normal Q-Q plot doesn’t look good. We can try log transforming the data and see if that helps.

## Anova Table (Type III tests)
## 
## Response: log_changePO4_uM
##                       Sum Sq Df  F value    Pr(>F)    
## (Intercept)           4.2420  1 353.0388 < 2.2e-16 ***
## temperature           0.2352  2   9.7859 0.0002137 ***
## treatment             1.3423  3  37.2365 1.239e-13 ***
## temperature:treatment 0.4092  6   5.6765 0.0001046 ***
## Residuals             0.7089 59                       
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

## Levene's Test for Homogeneity of Variance (center = median)
##       Df F value  Pr(>F)  
## group 11  2.5357 0.01079 *
##       59                  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

This is slightly better, but there is still quite a bit spread in our residuals which is making us fail the Levene test. Might need to think of a workaround here.

PO4 uptake/release - Pairwise comparisons

Pairwise comparsions for each level of temperature. Pretty much the only time we don’t see a significant different between treatments is at 8 degree between NP & P, 12 degrees between C & N and P & NP, and 16 degrees between C & N, and P & NP.

## $emmeans
## temperature = 8:
##  treatment    emmean         SE df  lower.CL  upper.CL
##  C         0.8408372 0.04475079 59 0.7512911 0.9303833
##  N         0.5230641 0.04475079 59 0.4335180 0.6126103
##  P         1.1541683 0.04902204 59 1.0560754 1.2522611
##  NP        1.0593662 0.04475079 59 0.9698201 1.1489123
## 
## temperature = 12:
##  treatment    emmean         SE df  lower.CL  upper.CL
##  C         0.8786424 0.04475079 59 0.7890962 0.9681885
##  N         0.8308817 0.04475079 59 0.7413356 0.9204278
##  P         1.3029583 0.04475079 59 1.2134122 1.3925044
##  NP        1.3115208 0.04475079 59 1.2219746 1.4010669
## 
## temperature = 16:
##  treatment    emmean         SE df  lower.CL  upper.CL
##  C         1.0999912 0.04475079 59 1.0104451 1.1895374
##  N         1.0999912 0.04475079 59 1.0104451 1.1895374
##  P         1.3029583 0.04475079 59 1.2134122 1.3925044
##  NP        1.3115208 0.04475079 59 1.2219746 1.4010669
## 
## Confidence level used: 0.95 
## 
## $contrasts
## temperature = 8:
##  contrast      estimate         SE df t.ratio p.value
##  C - N     3.177731e-01 0.06328718 59   5.021  <.0001
##  C - P    -3.133311e-01 0.06637615 59  -4.721  0.0001
##  C - NP   -2.185290e-01 0.06328718 59  -3.453  0.0055
##  N - P    -6.311041e-01 0.06637615 59  -9.508  <.0001
##  N - NP   -5.363021e-01 0.06328718 59  -8.474  <.0001
##  P - NP    9.480208e-02 0.06637615 59   1.428  0.4871
## 
## temperature = 12:
##  contrast      estimate         SE df t.ratio p.value
##  C - N     4.776069e-02 0.06328718 59   0.755  0.8743
##  C - P    -4.243159e-01 0.06328718 59  -6.705  <.0001
##  C - NP   -4.328784e-01 0.06328718 59  -6.840  <.0001
##  N - P    -4.720766e-01 0.06328718 59  -7.459  <.0001
##  N - NP   -4.806391e-01 0.06328718 59  -7.595  <.0001
##  P - NP   -8.562444e-03 0.06328718 59  -0.135  0.9991
## 
## temperature = 16:
##  contrast      estimate         SE df t.ratio p.value
##  C - N    -2.775558e-16 0.06328718 59   0.000  1.0000
##  C - P    -2.029671e-01 0.06328718 59  -3.207  0.0113
##  C - NP   -2.115295e-01 0.06328718 59  -3.342  0.0077
##  N - P    -2.029671e-01 0.06328718 59  -3.207  0.0113
##  N - NP   -2.115295e-01 0.06328718 59  -3.342  0.0077
##  P - NP   -8.562444e-03 0.06328718 59  -0.135  0.9991
## 
## P value adjustment: tukey method for comparing a family of 4 estimates

Pairwise comparsions for each level of nutrient treatment. In the control nutrient treatment, there is a different between 8 and 16 degrees and 12 and 16 degrees. In the nitrogen nutrient treatment, there is a statistically significant difference with temperature. In the P treatment, there is a statastically significant difference between 8 and 12 and 8 and 16, but not between 12 and 16. Finally, in the NP treatment, therre is a difference between 8 and 12 and 8 and 16, but not between 12 and 16.

## $emmeans
## treatment = C:
##  temperature    emmean         SE df  lower.CL  upper.CL
##  8           0.8408372 0.04475079 59 0.7512911 0.9303833
##  12          0.8786424 0.04475079 59 0.7890962 0.9681885
##  16          1.0999912 0.04475079 59 1.0104451 1.1895374
## 
## treatment = N:
##  temperature    emmean         SE df  lower.CL  upper.CL
##  8           0.5230641 0.04475079 59 0.4335180 0.6126103
##  12          0.8308817 0.04475079 59 0.7413356 0.9204278
##  16          1.0999912 0.04475079 59 1.0104451 1.1895374
## 
## treatment = P:
##  temperature    emmean         SE df  lower.CL  upper.CL
##  8           1.1541683 0.04902204 59 1.0560754 1.2522611
##  12          1.3029583 0.04475079 59 1.2134122 1.3925044
##  16          1.3029583 0.04475079 59 1.2134122 1.3925044
## 
## treatment = NP:
##  temperature    emmean         SE df  lower.CL  upper.CL
##  8           1.0593662 0.04475079 59 0.9698201 1.1489123
##  12          1.3115208 0.04475079 59 1.2219746 1.4010669
##  16          1.3115208 0.04475079 59 1.2219746 1.4010669
## 
## Confidence level used: 0.95 
## 
## $contrasts
## treatment = C:
##  contrast      estimate         SE df t.ratio p.value
##  8 - 12   -3.780519e-02 0.06328718 59  -0.597  0.8221
##  8 - 16   -2.591540e-01 0.06328718 59  -4.095  0.0004
##  12 - 16  -2.213489e-01 0.06328718 59  -3.498  0.0026
## 
## treatment = N:
##  contrast      estimate         SE df t.ratio p.value
##  8 - 12   -3.078176e-01 0.06328718 59  -4.864  <.0001
##  8 - 16   -5.769271e-01 0.06328718 59  -9.116  <.0001
##  12 - 16  -2.691095e-01 0.06328718 59  -4.252  0.0002
## 
## treatment = P:
##  contrast      estimate         SE df t.ratio p.value
##  8 - 12   -1.487901e-01 0.06637615 59  -2.242  0.0725
##  8 - 16   -1.487901e-01 0.06637615 59  -2.242  0.0725
##  12 - 16  -9.714451e-17 0.06328718 59   0.000  1.0000
## 
## treatment = NP:
##  contrast      estimate         SE df t.ratio p.value
##  8 - 12   -2.521546e-01 0.06328718 59  -3.984  0.0005
##  8 - 16   -2.521546e-01 0.06328718 59  -3.984  0.0005
##  12 - 16  -3.053113e-16 0.06328718 59   0.000  1.0000
## 
## P value adjustment: tukey method for comparing a family of 3 estimates

PO4 uptake/release - Two-way ANOVA with Robust Estimation

Tutorial found here: http://rcompanion.org/rcompanion/d_08a.html More info on non-parametric tests and R packages here: https://journal.r-project.org/archive/2016/RJ-2016-027/RJ-2016-027.pdf

First, Produce Huber M-estimators and confidence intervals by group NOT PRINTING AT THIS TIME

Data Visualization - Beaker Chla

Before we jump into any statistical analyses, just want to get an idea of how things play out by treatment and temperature .

Beaker Chla - Data table summary

Incubation experiment nutrient data summary: means, standard deviation, and number of replicates per treatment
tempxtrt treatment temperature meanphytochla sdphytochla n
8Control C 8 19.947 6.011 6
8N+P NP 8 71.782 5.824 6
8Nitrogen N 8 19.908 6.649 6
8Phosphorus P 8 27.330 5.930 5
12Control C 12 22.712 7.839 6
12N+P NP 12 65.117 5.632 6
12Nitrogen N 12 21.140 8.846 6
12Phosphorus P 12 17.715 6.384 6
16Control C 16 19.338 7.403 6
16N+P NP 16 42.807 12.355 6
16Nitrogen N 16 37.978 6.251 6
16Phosphorus P 16 19.847 4.565 6

Beaker Chla - Descriptive statistics and plots

Here we have some summary statistics for Beaker Chla. First we are looking at how much chlorophyll a there was in the beaker at the end of experiment by TEMPERTURE and by TREATMENT separately. I’ve also plotted boxplots with individual points. Not sure where all these phytoplankton came from, since we used filtered lake water.

## ------------------------------------------------------------------------- 
## chl_a_phyto ~ temperature
## 
## Summary: 
## n pairs: 71, valid: 71 (100.0%), missings: 0 (0.0%), groups: 3
## 
##                            
##             8     12     16
## mean     35.1   31.7   30.0
## median   25.4   22.9   28.7
## sd       23.2   20.9   13.2
## IQR      31.6   23.7   19.3
## n          23     24     24
## np      32.4%  33.8%  33.8%
## NAs         0      0      0
## 0s          0      0      0
## 
## Kruskal-Wallis rank sum test:
##   Kruskal-Wallis chi-squared = 0.37119, df = 2, p-value = 0.8306

## ------------------------------------------------------------------------- 
## chl_a_phyto ~ treatment
## 
## Summary: 
## n pairs: 71, valid: 71 (100.0%), missings: 0 (0.0%), groups: 4
## 
##                                   
##             C      N      P     NP
## mean     20.7   26.3   21.3   59.9
## median   21.8   28.0   22.0   64.0
## sd        6.9   10.9    6.7   15.1
## IQR       8.0   19.0    8.2   15.7
## n          18     18     17     18
## np      25.4%  25.4%  23.9%  25.4%
## NAs         0      0      0      0
## 0s          0      0      0      0
## 
## Kruskal-Wallis rank sum test:
##   Kruskal-Wallis chi-squared = 35.38, df = 3, p-value = 1.013e-07

Beaker Chla - ANOVA results - with intxn

From our preliminary glance at the data, it doesn’t look like there is much a difference in the means, although in the 8 degree treatments, the spread was quite wide. Comparing across nutrient treatments, NP has the highest mean chla with N as the runner up.

## Anova Table (Type III tests)
## 
## Response: chl_a_phyto
##                        Sum Sq Df F value    Pr(>F)    
## (Intercept)            2387.2  1  45.292 7.740e-09 ***
## temperature              38.8  2   0.368    0.6937    
## treatment             11137.6  3  70.436 < 2.2e-16 ***
## temperature:treatment  4042.5  6  12.783 3.445e-09 ***
## Residuals              3109.8 59                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Studying the output of the ANOVA table (Type III SS), we see there is a evidence of a significant interaction effect (F(6,59)=12.783, p<0.0001). The tests for the main effects of temperature is not significant, but the effect ot treatment is highly significant (F(3,59)=70.436, p<0.0001).

After fitting an ANOVA model it is important to always check the relevant model assumptions. This includes making QQ-plots and residual plots. Will use Levene’s test of equality of variances too.

## Levene's Test for Homogeneity of Variance (center = median)
##       Df F value Pr(>F)
## group 11  0.3043 0.9824
##       59

Data fit the model assumptions. We can move on.

Beaker Chla - Pairwise comparisons

Pairwise comparsions for each level of temperature. At 8 degrees, NP is different than C, N, and P. At 12 degrees, NP is different than C, N, and P. At 16 degrees, both N and NP are significantly different than the Control. N vs P and P versus NP are both significantly different from each other, but N and NP are not significantly different.

## $emmeans
## temperature = 8:
##  treatment   emmean       SE df lower.CL upper.CL
##  C         19.94667 2.963885 59 14.01595 25.87739
##  N         19.90833 2.963885 59 13.97761 25.83905
##  P         27.33000 3.246773 59 20.83322 33.82678
##  NP        71.78167 2.963885 59 65.85095 77.71239
## 
## temperature = 12:
##  treatment   emmean       SE df lower.CL upper.CL
##  C         22.71167 2.963885 59 16.78095 28.64239
##  N         21.14000 2.963885 59 15.20928 27.07072
##  P         17.71500 2.963885 59 11.78428 23.64572
##  NP        65.11667 2.963885 59 59.18595 71.04739
## 
## temperature = 16:
##  treatment   emmean       SE df lower.CL upper.CL
##  C         19.33833 2.963885 59 13.40761 25.26905
##  N         37.97833 2.963885 59 32.04761 43.90905
##  P         19.84667 2.963885 59 13.91595 25.77739
##  NP        42.80667 2.963885 59 36.87595 48.73739
## 
## Confidence level used: 0.95 
## 
## $contrasts
## temperature = 8:
##  contrast     estimate       SE df t.ratio p.value
##  C - N      0.03833333 4.191566 59   0.009  1.0000
##  C - P     -7.38333333 4.396151 59  -1.679  0.3435
##  C - NP   -51.83500000 4.191566 59 -12.367  <.0001
##  N - P     -7.42166667 4.396151 59  -1.688  0.3390
##  N - NP   -51.87333333 4.191566 59 -12.376  <.0001
##  P - NP   -44.45166667 4.396151 59 -10.111  <.0001
## 
## temperature = 12:
##  contrast     estimate       SE df t.ratio p.value
##  C - N      1.57166667 4.191566 59   0.375  0.9819
##  C - P      4.99666667 4.191566 59   1.192  0.6342
##  C - NP   -42.40500000 4.191566 59 -10.117  <.0001
##  N - P      3.42500000 4.191566 59   0.817  0.8461
##  N - NP   -43.97666667 4.191566 59 -10.492  <.0001
##  P - NP   -47.40166667 4.191566 59 -11.309  <.0001
## 
## temperature = 16:
##  contrast     estimate       SE df t.ratio p.value
##  C - N    -18.64000000 4.191566 59  -4.447  0.0002
##  C - P     -0.50833333 4.191566 59  -0.121  0.9994
##  C - NP   -23.46833333 4.191566 59  -5.599  <.0001
##  N - P     18.13166667 4.191566 59   4.326  0.0003
##  N - NP    -4.82833333 4.191566 59  -1.152  0.6591
##  P - NP   -22.96000000 4.191566 59  -5.478  <.0001
## 
## P value adjustment: tukey method for comparing a family of 4 estimates

Pairwise comparsions for each level of nutrient treatment. In the control nutrient treatment, there is a different between 8 and 16 degrees and 12 and 16 degrees. In the nitrogen nutrient treatment, there is a statistically significant difference with temperature. In the P treatment, there is a statastically significant difference between 8 and 12 and 8 and 16, but not between 12 and 16. Finally, in the NP treatment, therre is a difference between 8 and 12 and 8 and 16, but not between 12 and 16.

## $emmeans
## treatment = C:
##  temperature    emmean         SE df  lower.CL  upper.CL
##  8           0.8408372 0.04475079 59 0.7512911 0.9303833
##  12          0.8786424 0.04475079 59 0.7890962 0.9681885
##  16          1.0999912 0.04475079 59 1.0104451 1.1895374
## 
## treatment = N:
##  temperature    emmean         SE df  lower.CL  upper.CL
##  8           0.5230641 0.04475079 59 0.4335180 0.6126103
##  12          0.8308817 0.04475079 59 0.7413356 0.9204278
##  16          1.0999912 0.04475079 59 1.0104451 1.1895374
## 
## treatment = P:
##  temperature    emmean         SE df  lower.CL  upper.CL
##  8           1.1541683 0.04902204 59 1.0560754 1.2522611
##  12          1.3029583 0.04475079 59 1.2134122 1.3925044
##  16          1.3029583 0.04475079 59 1.2134122 1.3925044
## 
## treatment = NP:
##  temperature    emmean         SE df  lower.CL  upper.CL
##  8           1.0593662 0.04475079 59 0.9698201 1.1489123
##  12          1.3115208 0.04475079 59 1.2219746 1.4010669
##  16          1.3115208 0.04475079 59 1.2219746 1.4010669
## 
## Confidence level used: 0.95 
## 
## $contrasts
## treatment = C:
##  contrast      estimate         SE df t.ratio p.value
##  8 - 12   -3.780519e-02 0.06328718 59  -0.597  0.8221
##  8 - 16   -2.591540e-01 0.06328718 59  -4.095  0.0004
##  12 - 16  -2.213489e-01 0.06328718 59  -3.498  0.0026
## 
## treatment = N:
##  contrast      estimate         SE df t.ratio p.value
##  8 - 12   -3.078176e-01 0.06328718 59  -4.864  <.0001
##  8 - 16   -5.769271e-01 0.06328718 59  -9.116  <.0001
##  12 - 16  -2.691095e-01 0.06328718 59  -4.252  0.0002
## 
## treatment = P:
##  contrast      estimate         SE df t.ratio p.value
##  8 - 12   -1.487901e-01 0.06637615 59  -2.242  0.0725
##  8 - 16   -1.487901e-01 0.06637615 59  -2.242  0.0725
##  12 - 16  -9.714451e-17 0.06328718 59   0.000  1.0000
## 
## treatment = NP:
##  contrast      estimate         SE df t.ratio p.value
##  8 - 12   -2.521546e-01 0.06328718 59  -3.984  0.0005
##  8 - 16   -2.521546e-01 0.06328718 59  -3.984  0.0005
##  12 - 16  -3.053113e-16 0.06328718 59   0.000  1.0000
## 
## P value adjustment: tukey method for comparing a family of 3 estimates

NEP/ER/GPP Assays

Data Visualization - GPP (gross primary production)

Before we jump into any statistical analyses, just want to get an idea of how things play out by treatment and temperature .

GPP - Data table summary

Incubation experiment GPP data summary: means, standard deviation, and number of replicates per treatment
treatment temperature meanGPP sdGPP n
C 8 41.000 16.646 6
N 8 38.483 6.845 6
NP 8 45.567 19.708 6
P 8 51.250 6.149 6
C 12 60.517 17.748 6
N 12 38.167 6.124 6
NP 12 37.383 8.541 6
P 12 40.467 5.545 6
C 16 61.100 4.330 6
N 16 54.300 4.951 6
NP 16 58.217 6.065 6
P 16 52.483 3.459 6

GPP - Descriptive statistics and plots

Here we have some summary statistics for GPP. First we are looking at GPP by TEMPERTURE and by TREATMENT separately.

## ------------------------------------------------------------------------- 
## GPP_ugcm2h ~ temperature
## 
## Summary: 
## n pairs: 72, valid: 72 (100.0%), missings: 0 (0.0%), groups: 3
## 
##                            
##             8     12     16
## mean     44.1   44.1   56.5
## median   43.0   43.2   58.0
## sd       13.7   13.9    5.6
## IQR      20.1   11.4    9.6
## n          24     24     24
## np      33.3%  33.3%  33.3%
## NAs         0      0      0
## 0s          0      0      0
## 
## Kruskal-Wallis rank sum test:
##   Kruskal-Wallis chi-squared = 22.444, df = 2, p-value = 1.337e-05

## ------------------------------------------------------------------------- 
## GPP_ugcm2h ~ treatment
## 
## Summary: 
## n pairs: 72, valid: 72 (100.0%), missings: 0 (0.0%), groups: 4
## 
##                                   
##             C      N      P     NP
## mean     54.2   43.6   48.1   47.1
## median   56.8   42.8   49.3   46.8
## sd       16.5    9.6    7.4   15.0
## IQR      19.1   11.8    8.0   27.9
## n          18     18     18     18
## np      25.0%  25.0%  25.0%  25.0%
## NAs         0      0      0      0
## 0s          0      0      0      0
## 
## Kruskal-Wallis rank sum test:
##   Kruskal-Wallis chi-squared = 5.5192, df = 3, p-value = 0.1375

GPP - ANOVA results - with intxn

From our preliminary glance at the data, it looks like GPP might be higher on average in the warmest temperature treatment (16 degrees). As for the nutrients, it the controls have the widest spread in GPP and as a result have the highest mean.

## Anova Table (Type III tests)
## 
## Response: GPP_ugcm2h
##                        Sum Sq Df F value    Pr(>F)    
## (Intercept)           10086.0  1 93.2200 7.936e-14 ***
## temperature            1570.5  2  7.2577  0.001504 ** 
## treatment               566.6  3  1.7455  0.167331    
## temperature:treatment  1969.6  6  3.0340  0.011710 *  
## Residuals              6491.7 60                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Studying the output of the ANOVA table (Type III SS), we see there is a evidence of a significant interaction effect (F(6,60)=3.0340, p=0.011710). The tests for the main effects of treatment is not significant, but the effect of temperature is significant (F(2,60)=7.2577, p=0.001504).

After fitting an ANOVA model it is important to always check the relevant model assumptions. This includes making QQ-plots and residual plots. Will use Levene’s test of equality of variances too.

## Levene's Test for Homogeneity of Variance (center = median)
##       Df F value   Pr(>F)   
## group 11  2.5787 0.009477 **
##       60                    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Data doesn’t fit model assumptions, unfortunately. Let’s try a log transformation…

## Anova Table (Type III tests)
## 
## Response: log_GPP_ugcm2h
##                       Sum Sq Df  F value    Pr(>F)    
## (Intercept)           7.4772  1 108.8152 4.215e-15 ***
## temperature           0.8033  2   5.8450  0.004795 ** 
## treatment             0.4700  3   2.2800  0.088469 .  
## temperature:treatment 0.8508  6   2.0636  0.070960 .  
## Residuals             4.1229 60                       
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

## Levene's Test for Homogeneity of Variance (center = median)
##       Df F value    Pr(>F)    
## group 11  3.6164 0.0005946 ***
##       60                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## Anova Table (Type III tests)
## 
## Response: sqrt_GPP
##                        Sum Sq Df  F value    Pr(>F)    
## (Intercept)           238.241  1 423.1044 < 2.2e-16 ***
## temperature             8.570  2   7.6095  0.001134 ** 
## treatment               3.359  3   1.9887  0.125263    
## temperature:treatment  10.324  6   3.0557  0.011245 *  
## Residuals              33.785 60                       
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

## Levene's Test for Homogeneity of Variance (center = median)
##       Df F value   Pr(>F)   
## group 11  2.9748 0.003271 **
##       60                    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Neither log or square root transformations help. Will need to try something different. In the mean time, let’s look at some pairwise comparisons ANYWAY!

GPP - Pairwise comparisons

Pairwise comparsions for each level of temperature. At 8 degrees, there is no difference between treatments. At 12 degrees, the Control is different than all the other nutrient treatments. And at 16 degrees there is no different between treatments.

## $emmeans
## temperature = 8:
##  treatment   emmean       SE df lower.CL upper.CL
##  C         41.00000 4.246481 60 32.50577 49.49423
##  N         38.48333 4.246481 60 29.98911 46.97756
##  P         51.25000 4.246481 60 42.75577 59.74423
##  NP        45.56667 4.246481 60 37.07244 54.06089
## 
## temperature = 12:
##  treatment   emmean       SE df lower.CL upper.CL
##  C         60.51667 4.246481 60 52.02244 69.01089
##  N         38.16667 4.246481 60 29.67244 46.66089
##  P         40.46667 4.246481 60 31.97244 48.96089
##  NP        37.38333 4.246481 60 28.88911 45.87756
## 
## temperature = 16:
##  treatment   emmean       SE df lower.CL upper.CL
##  C         61.10000 4.246481 60 52.60577 69.59423
##  N         54.30000 4.246481 60 45.80577 62.79423
##  P         52.48333 4.246481 60 43.98911 60.97756
##  NP        58.21667 4.246481 60 49.72244 66.71089
## 
## Confidence level used: 0.95 
## 
## $contrasts
## temperature = 8:
##  contrast    estimate       SE df t.ratio p.value
##  C - N      2.5166667 6.005431 60   0.419  0.9750
##  C - P    -10.2500000 6.005431 60  -1.707  0.3292
##  C - NP    -4.5666667 6.005431 60  -0.760  0.8718
##  N - P    -12.7666667 6.005431 60  -2.126  0.1567
##  N - NP    -7.0833333 6.005431 60  -1.179  0.6420
##  P - NP     5.6833333 6.005431 60   0.946  0.7800
## 
## temperature = 12:
##  contrast    estimate       SE df t.ratio p.value
##  C - N     22.3500000 6.005431 60   3.722  0.0024
##  C - P     20.0500000 6.005431 60   3.339  0.0077
##  C - NP    23.1333333 6.005431 60   3.852  0.0016
##  N - P     -2.3000000 6.005431 60  -0.383  0.9807
##  N - NP     0.7833333 6.005431 60   0.130  0.9992
##  P - NP     3.0833333 6.005431 60   0.513  0.9556
## 
## temperature = 16:
##  contrast    estimate       SE df t.ratio p.value
##  C - N      6.8000000 6.005431 60   1.132  0.6712
##  C - P      8.6166667 6.005431 60   1.435  0.4830
##  C - NP     2.8833333 6.005431 60   0.480  0.9632
##  N - P      1.8166667 6.005431 60   0.303  0.9903
##  N - NP    -3.9166667 6.005431 60  -0.652  0.9144
##  P - NP    -5.7333333 6.005431 60  -0.955  0.7754
## 
## P value adjustment: tukey method for comparing a family of 4 estimates

Pairwise comparsions for each level of nutrient treatment. In the control treatment, there is a difference between 8 and 12 and 8 and 16, but none between 12 and 16. In the nitrogen treatment, there is a different between 8 and 16 ad 12 and 16, but not between 8 and 12. In the P treatment, there are no differences between temperatures. In the NP treatment, there is a difference between 8 and 16 and 12 and 16 but none between 8 and 12.

## $emmeans
## treatment = C:
##  temperature   emmean       SE df lower.CL upper.CL
##  8           41.00000 4.246481 60 32.50577 49.49423
##  12          60.51667 4.246481 60 52.02244 69.01089
##  16          61.10000 4.246481 60 52.60577 69.59423
## 
## treatment = N:
##  temperature   emmean       SE df lower.CL upper.CL
##  8           38.48333 4.246481 60 29.98911 46.97756
##  12          38.16667 4.246481 60 29.67244 46.66089
##  16          54.30000 4.246481 60 45.80577 62.79423
## 
## treatment = P:
##  temperature   emmean       SE df lower.CL upper.CL
##  8           51.25000 4.246481 60 42.75577 59.74423
##  12          40.46667 4.246481 60 31.97244 48.96089
##  16          52.48333 4.246481 60 43.98911 60.97756
## 
## treatment = NP:
##  temperature   emmean       SE df lower.CL upper.CL
##  8           45.56667 4.246481 60 37.07244 54.06089
##  12          37.38333 4.246481 60 28.88911 45.87756
##  16          58.21667 4.246481 60 49.72244 66.71089
## 
## Confidence level used: 0.95 
## 
## $contrasts
## treatment = C:
##  contrast    estimate       SE df t.ratio p.value
##  8 - 12   -19.5166667 6.005431 60  -3.250  0.0053
##  8 - 16   -20.1000000 6.005431 60  -3.347  0.0040
##  12 - 16   -0.5833333 6.005431 60  -0.097  0.9948
## 
## treatment = N:
##  contrast    estimate       SE df t.ratio p.value
##  8 - 12     0.3166667 6.005431 60   0.053  0.9985
##  8 - 16   -15.8166667 6.005431 60  -2.634  0.0285
##  12 - 16  -16.1333333 6.005431 60  -2.686  0.0249
## 
## treatment = P:
##  contrast    estimate       SE df t.ratio p.value
##  8 - 12    10.7833333 6.005431 60   1.796  0.1797
##  8 - 16    -1.2333333 6.005431 60  -0.205  0.9770
##  12 - 16  -12.0166667 6.005431 60  -2.001  0.1207
## 
## treatment = NP:
##  contrast    estimate       SE df t.ratio p.value
##  8 - 12     8.1833333 6.005431 60   1.363  0.3668
##  8 - 16   -12.6500000 6.005431 60  -2.106  0.0971
##  12 - 16  -20.8333333 6.005431 60  -3.469  0.0028
## 
## P value adjustment: tukey method for comparing a family of 3 estimates

Data Visualization - ER (microbial respiration)

Before we jump into any statistical analyses, just want to get an idea of how things play out by treatment and temperature .

ER - Data table summary

Incubation experiment ER data summary: means, standard deviation, and number of replicates per treatment
treatment temperature meanER sdER n
C 8 0.517 4.449 6
N 8 -13.533 5.007 6
NP 8 -21.333 10.945 6
P 8 -13.983 3.071 6
C 12 -26.450 12.501 6
N 12 -14.333 4.717 6
NP 12 -10.783 5.969 6
P 12 -18.783 6.492 6
C 16 -42.067 2.196 6
N 16 -35.217 4.990 6
NP 16 -39.783 6.218 6
P 16 -42.583 5.312 6

ER - Descriptive statistics and plots

Here we have some summary statistics for ER. First we are looking at ER by TEMPERTURE and by TREATMENT separately.

## ------------------------------------------------------------------------- 
## ER_ugcm2h ~ temperature
## 
## Summary: 
## n pairs: 72, valid: 72 (100.0%), missings: 0 (0.0%), groups: 3
## 
##                            
##             8     12     16
## mean    -12.1  -17.6  -39.9
## median  -12.8  -15.2  -40.8
## sd       10.2    9.6    5.5
## IQR       9.2    9.6    6.6
## n          24     24     24
## np      33.3%  33.3%  33.3%
## NAs         0      0      0
## 0s          0      0      0
## 
## Kruskal-Wallis rank sum test:
##   Kruskal-Wallis chi-squared = 43.121, df = 2, p-value = 4.328e-10

## ------------------------------------------------------------------------- 
## ER_ugcm2h ~ treatment
## 
## Summary: 
## n pairs: 72, valid: 72 (100.0%), missings: 0 (0.0%), groups: 4
## 
##                                   
##             C      N      P     NP
## mean    -22.7  -21.0  -25.1  -24.0
## median  -24.6  -16.9  -20.5  -17.8
## sd       19.5   11.3   13.7   14.5
## IQR      37.7   19.6   23.9   23.7
## n          18     18     18     18
## np      25.0%  25.0%  25.0%  25.0%
## NAs         0      0      0      0
## 0s          0      0      0      0
## 
## Kruskal-Wallis rank sum test:
##   Kruskal-Wallis chi-squared = 0.83723, df = 3, p-value = 0.8405

ER - ANOVA results - with intxn

From our preliminary glance at the data, it looks like ER might be higher on average in the warmest temperature treatment (16 degrees). As for the nutrients, the control has the widest spread in ER, but the highest mean ER is in the N treatment.

## Anova Table (Type III tests)
## 
## Response: ER_ugcm2h
##                       Sum Sq Df F value    Pr(>F)    
## (Intercept)              1.6  1  0.0365    0.8492    
## temperature           5568.8  2 63.4070 1.593e-15 ***
## treatment             1500.2  3 11.3876 5.202e-06 ***
## temperature:treatment 2357.7  6  8.9481 5.598e-07 ***
## Residuals             2634.8 60                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Studying the output of the ANOVA table (Type III SS), we see there is a evidence of a significant interaction effect (F(6,60)=8.9481, p<0.0001). The tests for the main effects of temperature treatment is significant (F(2,60)=63.4, p<0.00001) and so is nutrient treatment (F(3,60)=11.3876, p<0.0001.)

After fitting an ANOVA model it is important to always check the relevant model assumptions. This includes making QQ-plots and residual plots. Will use Levene’s test of equality of variances too.

## Levene's Test for Homogeneity of Variance (center = median)
##       Df F value  Pr(>F)  
## group 11  1.8059 0.07288 .
##       60                  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Looks like we meet the assumptions of normality so we can move onto pairwise comparisons.

ER - Pairwise comparisons

Pairwise comparsions for each level of temperature. At 8 degrees, there is a difference between the control and all nutrient treatments, but not between nutrient treatments. At 12 degrees, there is a different between C - N and C - NP. At 16 degrees, there is no statistically significant relationship between nutrient treatments.

## $emmeans
## temperature = 8:
##  treatment      emmean       SE df   lower.CL   upper.CL
##  C           0.5166667 2.705348 60  -4.894835   5.928168
##  N         -13.5333333 2.705348 60 -18.944835  -8.121832
##  P         -13.9833333 2.705348 60 -19.394835  -8.571832
##  NP        -21.3333333 2.705348 60 -26.744835 -15.921832
## 
## temperature = 12:
##  treatment      emmean       SE df   lower.CL   upper.CL
##  C         -26.4500000 2.705348 60 -31.861502 -21.038498
##  N         -14.3333333 2.705348 60 -19.744835  -8.921832
##  P         -18.7833333 2.705348 60 -24.194835 -13.371832
##  NP        -10.7833333 2.705348 60 -16.194835  -5.371832
## 
## temperature = 16:
##  treatment      emmean       SE df   lower.CL   upper.CL
##  C         -42.0666667 2.705348 60 -47.478168 -36.655165
##  N         -35.2166667 2.705348 60 -40.628168 -29.805165
##  P         -42.5833333 2.705348 60 -47.994835 -37.171832
##  NP        -39.7833333 2.705348 60 -45.194835 -34.371832
## 
## Confidence level used: 0.95 
## 
## $contrasts
## temperature = 8:
##  contrast    estimate      SE df t.ratio p.value
##  C - N     14.0500000 3.82594 60   3.672  0.0028
##  C - P     14.5000000 3.82594 60   3.790  0.0019
##  C - NP    21.8500000 3.82594 60   5.711  <.0001
##  N - P      0.4500000 3.82594 60   0.118  0.9994
##  N - NP     7.8000000 3.82594 60   2.039  0.1855
##  P - NP     7.3500000 3.82594 60   1.921  0.2302
## 
## temperature = 12:
##  contrast    estimate      SE df t.ratio p.value
##  C - N    -12.1166667 3.82594 60  -3.167  0.0126
##  C - P     -7.6666667 3.82594 60  -2.004  0.1980
##  C - NP   -15.6666667 3.82594 60  -4.095  0.0007
##  N - P      4.4500000 3.82594 60   1.163  0.6522
##  N - NP    -3.5500000 3.82594 60  -0.928  0.7900
##  P - NP    -8.0000000 3.82594 60  -2.091  0.1678
## 
## temperature = 16:
##  contrast    estimate      SE df t.ratio p.value
##  C - N     -6.8500000 3.82594 60  -1.790  0.2880
##  C - P      0.5166667 3.82594 60   0.135  0.9991
##  C - NP    -2.2833333 3.82594 60  -0.597  0.9327
##  N - P      7.3666667 3.82594 60   1.925  0.2285
##  N - NP     4.5666667 3.82594 60   1.194  0.6332
##  P - NP    -2.8000000 3.82594 60  -0.732  0.8839
## 
## P value adjustment: tukey method for comparing a family of 4 estimates

Pairwise comparsions for each level of nutrient treatment. In the control treatment, there is a difference between 8 - 12, 8 - 16, and between 12 - 16. In the N treatment, there is no difference between 8 and 12, but there is a difference between 8 - 16 and 12 - 16. In the P treatment, there is a difference between 8 - 16 and 12- 16. Lastly, in the NP treatment, there is a difference between 8 - 12, 8 - 16, and 12 - 16.

## $emmeans
## treatment = C:
##  temperature      emmean       SE df   lower.CL   upper.CL
##  8             0.5166667 2.705348 60  -4.894835   5.928168
##  12          -26.4500000 2.705348 60 -31.861502 -21.038498
##  16          -42.0666667 2.705348 60 -47.478168 -36.655165
## 
## treatment = N:
##  temperature      emmean       SE df   lower.CL   upper.CL
##  8           -13.5333333 2.705348 60 -18.944835  -8.121832
##  12          -14.3333333 2.705348 60 -19.744835  -8.921832
##  16          -35.2166667 2.705348 60 -40.628168 -29.805165
## 
## treatment = P:
##  temperature      emmean       SE df   lower.CL   upper.CL
##  8           -13.9833333 2.705348 60 -19.394835  -8.571832
##  12          -18.7833333 2.705348 60 -24.194835 -13.371832
##  16          -42.5833333 2.705348 60 -47.994835 -37.171832
## 
## treatment = NP:
##  temperature      emmean       SE df   lower.CL   upper.CL
##  8           -21.3333333 2.705348 60 -26.744835 -15.921832
##  12          -10.7833333 2.705348 60 -16.194835  -5.371832
##  16          -39.7833333 2.705348 60 -45.194835 -34.371832
## 
## Confidence level used: 0.95 
## 
## $contrasts
## treatment = C:
##  contrast  estimate      SE df t.ratio p.value
##  8 - 12    26.96667 3.82594 60   7.048  <.0001
##  8 - 16    42.58333 3.82594 60  11.130  <.0001
##  12 - 16   15.61667 3.82594 60   4.082  0.0004
## 
## treatment = N:
##  contrast  estimate      SE df t.ratio p.value
##  8 - 12     0.80000 3.82594 60   0.209  0.9762
##  8 - 16    21.68333 3.82594 60   5.667  <.0001
##  12 - 16   20.88333 3.82594 60   5.458  <.0001
## 
## treatment = P:
##  contrast  estimate      SE df t.ratio p.value
##  8 - 12     4.80000 3.82594 60   1.255  0.4263
##  8 - 16    28.60000 3.82594 60   7.475  <.0001
##  12 - 16   23.80000 3.82594 60   6.221  <.0001
## 
## treatment = NP:
##  contrast  estimate      SE df t.ratio p.value
##  8 - 12   -10.55000 3.82594 60  -2.757  0.0208
##  8 - 16    18.45000 3.82594 60   4.822  <.0001
##  12 - 16   29.00000 3.82594 60   7.580  <.0001
## 
## P value adjustment: tukey method for comparing a family of 3 estimates

Pigment Results

Data Visualization - Pigments (Chla A)

Chl A - Data table summary

The main thing to notice here compared to the other data is that our sample size is small (n=3 per trt combo)
Incubation experiment pigment data summary: means, standard deviation, and number of replicates per treatment
pigment_ID treatment temperature mean_ug_cm2 sd_mass_ug_cm2 n
chla C 12 2.422 1.358 3
chla N 12 1.370 1.690 3
chla NP 12 1.425 0.572 3
chla P 12 2.688 1.682 3
chla C 16 2.440 0.744 3
chla N 16 2.866 0.815 3
chla NP 16 1.926 0.662 3
chla P 16 1.801 1.386 3

Chl A - Descriptive statistics and plots

## ------------------------------------------------------------------------- 
## chla ~ temperature
## 
## Summary: 
## n pairs: 24, valid: 24 (100.0%), missings: 0 (0.0%), groups: 2
## 
##                     
##            12     16
## mean      2.0    2.3
## median    1.7    2.3
## sd        1.3    0.9
## IQR       1.5    1.1
## n          12     12
## np      50.0%  50.0%
## NAs         0      0
## 0s          0      0
## 
## Kruskal-Wallis rank sum test:
##   Kruskal-Wallis chi-squared = 0.75, df = 1, p-value = 0.3865

## ------------------------------------------------------------------------- 
## chla ~ treatment
## 
## Summary: 
## n pairs: 24, valid: 24 (100.0%), missings: 0 (0.0%), groups: 4
## 
##                                   
##             C      N     NP      P
## mean      2.4    2.1    1.7    2.2
## median    2.2    2.4    1.6    1.9
## sd        1.0    1.4    0.6    1.5
## IQR       1.1    2.1    0.7    1.9
## n           6      6      6      6
## np      25.0%  25.0%  25.0%  25.0%
## NAs         0      0      0      0
## 0s          0      0      0      0
## 
## Kruskal-Wallis rank sum test:
##   Kruskal-Wallis chi-squared = 1.0133, df = 3, p-value = 0.798

Chl A - ANOVA results - with intxn

## Anova Table (Type III tests)
## 
## Response: chla
##                        Sum Sq Df F value   Pr(>F)   
## (Intercept)           17.6031  1 12.3167 0.002905 **
## temperature            0.0004  1  0.0003 0.986164   
## treatment              4.1320  3  0.9637 0.433899   
## temperature:treatment  4.4382  3  1.0351 0.403617   
## Residuals             22.8673 16                    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Studying the output of the ANOVA table (Type III SS), <> it only thing that is significant is the intercept…whatever that means.

## Levene's Test for Homogeneity of Variance (center = median)
##       Df F value Pr(>F)
## group  7  0.3228 0.9325
##       16

Looks like we meet the assumptions of normality, but no need to move onto pairwise comparisons.

Data Visualization - Pigments (Chl B)

Chl B - Data table summary

The main thing to notice here compared to the other data is that our sample size is small (n=3 per trt combo)
Incubation experiment pigment data summary: means, standard deviation, and number of replicates per treatment
pigment_ID treatment temperature mean_ug_cm2 sd_mass_ug_cm2 n
chlb C 12 0.933 0.488 3
chlb N 12 0.573 0.654 3
chlb NP 12 0.656 0.294 3
chlb P 12 1.074 0.588 3
chlb C 16 1.004 0.244 3
chlb N 16 1.007 0.148 3
chlb NP 16 0.912 0.225 3
chlb P 16 0.499 0.248 3

Chl B - Descriptive statistics and plots

## ------------------------------------------------------------------------- 
## chlb ~ temperature
## 
## Summary: 
## n pairs: 24, valid: 24 (100.0%), missings: 0 (0.0%), groups: 2
## 
##                     
##            12     16
## mean      0.8    0.9
## median    0.8    0.9
## sd        0.5    0.3
## IQR       0.6    0.3
## n          12     12
## np      50.0%  50.0%
## NAs         0      0
## 0s          0      0
## 
## Kruskal-Wallis rank sum test:
##   Kruskal-Wallis chi-squared = 0.40333, df = 1, p-value = 0.5254

## ------------------------------------------------------------------------- 
## chlb ~ treatment
## 
## Summary: 
## n pairs: 24, valid: 24 (100.0%), missings: 0 (0.0%), groups: 4
## 
##                                   
##             C      N     NP      P
## mean      1.0    0.8    0.8    0.8
## median    0.9    0.9    0.8    0.7
## sd        0.3    0.5    0.3    0.5
## IQR       0.3    0.7    0.3    0.4
## n           6      6      6      6
## np      25.0%  25.0%  25.0%  25.0%
## NAs         0      0      0      0
## 0s          0      0      0      0
## 
## Kruskal-Wallis rank sum test:
##   Kruskal-Wallis chi-squared = 0.9, df = 3, p-value = 0.8254

Chl B - ANOVA results - with intxn

## Anova Table (Type III tests)
## 
## Response: chlb
##                        Sum Sq Df F value    Pr(>F)    
## (Intercept)           2.61308  1 16.1806 0.0009843 ***
## temperature           0.00753  1  0.0466 0.8317406    
## treatment             0.49485  3  1.0214 0.4092645    
## temperature:treatment 0.87158  3  1.7990 0.1879399    
## Residuals             2.58391 16                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Studying the output of the ANOVA table (Type III SS), <> nothing is significantly different.

## Levene's Test for Homogeneity of Variance (center = median)
##       Df F value Pr(>F)
## group  7  0.4584 0.8503
##       16

Looks like we meet the assumptions of normality, but no need to move onto pairwise comparisons.

Data Visualization - Percent Green (Chlb/Chla)

Percent Green - Data table summary

The main thing to notice here compared to the other data is that our sample size is small (n=3 per trt combo)
Incubation experiment pigment data summary: means, standard deviation, and number of replicates per treatment
pigment_ID treatment temperature mean_ug_cm2 sd_mass_ug_cm2 n
perc_green C 12 39.467 3.370 3
perc_green N 12 45.440 7.199 3
perc_green NP 12 45.235 6.726 3
perc_green P 12 42.339 7.957 3
perc_green C 16 41.803 4.855 3
perc_green N 16 35.982 4.384 3
perc_green NP 16 48.411 6.013 3
perc_green P 16 31.972 8.403 3

Percent Green - Descriptive statistics and plots

## ------------------------------------------------------------------------- 
## perc_green ~ temperature
## 
## Summary: 
## n pairs: 24, valid: 24 (100.0%), missings: 0 (0.0%), groups: 2
## 
##                     
##            12     16
## mean     43.1   39.5
## median   41.4   39.0
## sd        6.1    8.3
## IQR       9.1    6.2
## n          12     12
## np      50.0%  50.0%
## NAs         0      0
## 0s          0      0
## 
## Kruskal-Wallis rank sum test:
##   Kruskal-Wallis chi-squared = 0.56333, df = 1, p-value = 0.4529

## ------------------------------------------------------------------------- 
## perc_green ~ treatment
## 
## Summary: 
## n pairs: 24, valid: 24 (100.0%), missings: 0 (0.0%), groups: 4
## 
##                                   
##             C      N     NP      P
## mean     40.6   40.7   46.8   37.2
## median   39.0   39.3   47.8   37.8
## sd        4.0    7.4    6.0    9.3
## IQR       4.2    3.8    8.3    5.1
## n           6      6      6      6
## np      25.0%  25.0%  25.0%  25.0%
## NAs         0      0      0      0
## 0s          0      0      0      0
## 
## Kruskal-Wallis rank sum test:
##   Kruskal-Wallis chi-squared = 5.4667, df = 3, p-value = 0.1406

Chl B - ANOVA results - with intxn

## Anova Table (Type III tests)
## 
## Response: perc_green
##                       Sum Sq Df  F value   Pr(>F)    
## (Intercept)           4672.9  1 116.3567 9.48e-09 ***
## temperature              8.2  1   0.2038   0.6577    
## treatment               71.4  3   0.5928   0.6287    
## temperature:treatment  241.9  3   2.0076   0.1535    
## Residuals              642.6 16                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Studying the output of the ANOVA table (Type III SS), <> nothing is significantly different.

## Levene's Test for Homogeneity of Variance (center = median)
##       Df F value Pr(>F)
## group  7  0.1917  0.983
##       16

Looks like we meet the assumptions of normality, but no need to move onto pairwise comparisons.

Correlations between Nutrients & Functional Responses

## Observations: 72
## Variables: 17
## $ treatment     <chr> "C", "C", "C", "C", "C", "C", "N", "N", "N", "N"...
## $ temperature   <chr> "8", "8", "8", "8", "8", "8", "8", "8", "8", "8"...
## $ sample_id     <chr> "8C-1", "8C-2", "8C-3", "8C-4", "8C-5", "8C-6", ...
## $ rep           <int> 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, ...
## $ NEP_ugcm2h    <dbl> 65.9, 28.8, 24.1, 26.6, 26.2, 54.0, 31.3, 29.4, ...
## $ ER_ugcm2h     <dbl> 0.1, 6.5, 5.2, -0.9, -4.0, -3.8, -15.5, -9.7, -1...
## $ GPP_ugcm2h    <dbl> 66.1, 35.3, 29.3, 27.4, 30.1, 57.8, 46.8, 39.0, ...
## $ changePO4_uM  <dbl> -6.681, -6.078, -5.923, -6.597, -1.115, -6.336, ...
## $ changeTDN_uM  <dbl> 5.887, 5.967, 9.587, 10.057, 6.307, 5.417, 20.59...
## $ changeDOC_mgL <dbl> -3.281, -3.241, -2.091, -2.781, -3.271, -1.891, ...
## $ changeNH4_mgL <dbl> -0.032, -0.022, 0.018, 0.018, -0.012, -0.012, -0...
## $ percTDNuptake <dbl> 32.54824, 32.99055, 53.00492, 55.60347, 34.87035...
## $ percPO4uptake <dbl> -835.125000, -759.750000, -740.375000, -824.6250...
## $ chl_a_phyto   <dbl> 23.52, 11.44, 23.30, 25.36, 22.91, 13.15, 14.74,...
## $ chla          <dbl> NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, ...
## $ chlb          <dbl> NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, ...
## $ perc_green    <dbl> NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, ...

Subplots from correlation matrices

##       cor 
## 0.3608053

##        cor 
## -0.3410461

##       cor 
## 0.3945505

##        cor 
## -0.3761142

##        cor 
## -0.2738877

##       cor 
## 0.2036755

##       cor 
## 0.3408052

##       cor 
## 0.8316538

##       cor 
## 0.2653822

##      cor 
## 0.619436